Problem: $ f(x)=\sum_{n=0}^{\infty }(-1)^n\dfrac{n+1}{ 5^n}x^n$ $\int_{0}^{1}f(x)dx=$
First, perform the requested integration. $\begin{aligned} &\phantom{=}\int\limits_{0}^{1}{f\left( x \right)dx} \\\\ &=\int_0^1\sum_{n=0}^{\infty }(-1)^n\dfrac{n+1}{ 5^n}x^ndx \\\\ &=\sum_{n=0}^{\infty }\int_0^1(-1)^n\dfrac{n+1}{ 5^n}x^ndx \\\\ &=\left. \sum_{n=0}^{\infty }(-1)^n\dfrac{n+1}{ 5^n}\dfrac{x^{n+1}}{n+1} \right|_{0}^{1} \\\\ &=\sum_{n=0}^{\infty }{\dfrac{(-1)^n}{{{5}^n}}} \end{aligned}$ Now recognize that the result is a geometric series with first term $1$ and common ratio $-\dfrac{1}{5}$. Hence, the series converges to $\dfrac{1}{1+\dfrac{1}{5}}=\dfrac{5}{6}$.